Kate baked some cookies. She put 12 cookies on each cookie sheet. If she baked 24 cookie sheets of cookies, how many cookies did she bake in all?

Answer:

a) Null hypothesis: [tex]\mu \geq 1936[/tex]

Alternative hypothesis: [tex]\mu <1936[/tex]

b) [tex] t = \frac{1736-1936}{\frac{635}{\sqrt{46}}}= -2.136[/tex]

The degrees of freedom are:

[tex] df = n-1= 46-1=45[/tex]

The p value for this case since is a left tailed test is given by:

[tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is lower than the significance level 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 1736

c) [tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is higher than the significance level 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly lower than 1736

d) For this case the answer is yes since when we change the significance level from 0.05 to 0.01 we see that the final decision changes.

Step-by-step explanation:

Part a

We are trying to proof the following system of hypothesis:

Null hypothesis: [tex]\mu \geq 1936[/tex]

Alternative hypothesis: [tex]\mu <1936[/tex]

Part b

We have the following data given:

[tex]\bar X =1736[/tex] minute s = 635; n = 46

And the statistic for this case is given by:

[tex] t = \frac{\bar X- \mu}{\frac{s}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] t = \frac{1736-1936}{\frac{635}{\sqrt{46}}}= -2.136[/tex]

The degrees of freedom are:

[tex] df = n-1= 46-1=45[/tex]

The p value for this case since is a left tailed test is given by:

[tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is lower than the significance level 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 1736

Part c

We have the same statistic t = -2.136

[tex] p_v = P(t_{45} <-2.136) = 0.0191[/tex]

And since the p value is higher than the significance level 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly lower than 1736

Part d

For this case the answer is yes since when we change the significance level from 0.05 to 0.01 we see that the final decision changes.

Null and alternative hypotheses are created for a t-test regarding trip times. The t-test is conducted at two different significance levels (α = 0.05 and α = 0.01). The conclusion may change with adjustments to α.

The hypotheses for this statistical test would be:

To test these hypotheses, we can perform a two-sided t-test. We are given:

For α = 0.05, if the t-statistic's associated p-value is lesser than α, we reject the null hypothesis. Likewise, for α = 0.01, the same process is undertaken.

The conclusion reached will depend on the calculated p-value. If the p-value is less than the chosen α, we will reject the null hypothesis and conclude that the drivers are not following safety regulations. If the p-value is greater than α, we fail to reject the null hypothesis and cannot conclude that drivers are not following safety regulations.

The test conclusion may change with changes in α, as α establishes the threshold for rejecting the null hypothesis.

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Answer:

Type I error: Concluding μ ≠ 40, when in fact μ = 40.

Type II error: Concluding μ = 40, when in fact μ ≠ 40.

Step-by-step explanation:

In this case we need to determine whether the mean amperage at which the 40-amp fuses burn out is 40.

The hypothesis to test this can be defined as follows:

H₀: The mean amperage at which the 40-amp fuses burn out is 40, i.e. μ = 40.

Hₐ: The mean amperage at which the 40-amp fuses burn out is different from 40, i.e. μ ≠ 40.

A type I error occurs when we discard a true null hypothesis (H₀) and a type II error is made when we fail to discard a false null hypothesis (H₀).

In this context, a type I error will be committed if we conclude that the mean amperage at which the 40-amp fuses burn out is different from 40, when in fact it is 40.

And a type II error will be committed if we conclude that the mean amperage at which the 40-amp fuses burn out is 40, when in fact it is different from 40.

The null hypothesis for the manufacturer of 40-amp fuses is that the mean amperage at which fuses burn out is 40 amps, while the alternative hypothesis is that the mean is not 40 amps. A Type I error is incorrectly rejecting a true null hypothesis, and a Type II error is failing to reject a false null hypothesis, both of which have consequences for the manufacturer in terms of production and safety.

A manufacturer of 40-amp fuses is interested in ensuring the mean amperage at which its fuses burn out is indeed 40 amps. To validate this, a sample of fuses must be tested, and a hypothesis test applied to the results. The null hypothesis (H0) of interest would state that the mean amperage at which the fuses burn out is 40 amps, formulated as H0: μ = 40, where μ is the population mean. The alternative hypothesis (H1) would indicate that the mean amperage is not 40 amps: H1: μ ≠ 40.

In this scenario, a Type I error would occur if the hypothesis test incorrectly rejects the null hypothesis when in fact the fuses do burn out at the mean of 40 amps. This could result in unnecessary production changes and costs for the manufacturer. Alternatively, a Type II error would occur if the test fails to reject the null hypothesis when the true mean amperage at which the fuses burn out is actually different from 40 amps. In such a case, the manufacturer might continue producing fuses that could either require frequent replacement or pose a risk of damage to electrical systems.

The determination of the true mean amperage is relevant because of the role of fuses and circuit breakers in protecting appliances and residents from harm due to large currents and because they are designed to tolerate high currents for brief periods, or in some cases like electric motors, for a longer duration. Thus, ensuring fuses operate correctly at their intended amperage is crucial for safety and functionality.

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Answer:

j / 9 = 5

j = 45

Step-by-step explanation:

j / 9 = 5

Multiply each side by 9

j/9 * 9 = 5*9

j = 45

Answer:

(A)29 cubic inch

Step-by-step explanation:

Diameter of the Cone =4 Inches

Height of the Cone =7 Inches

Volume of a Cone [tex]=\frac{1}{3}\pi r^2h[/tex]

First, we determine the radius, r.

Radius=Diameter/2=4/2=2 Inches

Therefore:

Volume of one paper cone [tex]=\frac{1}{3}\pi *2^2*7[/tex]

=29.32 cubic inch

=29 cubic inch (to the nearest cubic inch.)

Final answer:

The number 7,790,200 has zero decimal places as it is a whole number with no fractional part and ends in the unit's place.

Explanation:

The student asked how many decimal places are in 7,790,200. This question is related to place value and decimals in mathematics. The number 7,790,200 has no decimal part since it is a whole number and ends in the unit's place. Therefore, it contains zero decimal places. To expand on place value, when expressing numbers in decimal form, the position of a digit represents its value in powers of ten. For instance, the number 1837 can be decomposed as (1 × 10³) + (8 × 10²) + (3 × 10¹) + (7 × 10⁰), which shows the ones, tens, hundreds, and thousands places respectively.

Answer:

do you have a picture of the line graph ?

Answer is 4,000

Step-by-step explanation:

Answer:

So, if it grows by 6.75%, each year the population is 106.75% of the year before.

After 1 year, 370,000(1.0675). After two years, 370,000(1.0675)(1.0675).

370,000(1.0675)12 = your answer

Step-by-step explanation:

Answer:

asdfgbhnjm

Step-by-step explanation:

zxcvb

We have been given that a baseball player threw 82 strikes out of 103 pitches. We are asked to find the percentage of pitches that were strikes.

To solve our given problem, we need to find strikes are what percent of pitches.

[tex]\text{Percentage of pitches that were strikes}=\frac{82}{103}\times 100\%[/tex]

[tex]\text{Percentage of pitches that were strikes}=0.7961165\times 100\%[/tex]

[tex]\text{Percentage of pitches that were strikes}=79.61165\%[/tex]

[tex]\text{Percentage of pitches that were strikes}\approx 79.6\%[/tex]

Therefore, approximately [tex]79.6\%[/tex] of pitches were strikes.

Answer:

0.225

Step-by-step explanation:

Total outcomes of choosing 5 out of 18 members = 18C5

Outcomes of choosing 2 out 11 favourers, 3 out of 7 members  =          11C2 & 7C3

Probability = Favourable outcomes / Total outcomes

= ( 11C2 x 7C3 ) / 18C5

[ { 11 ! / 2! 9! } {7 ! / 3! 4! } ]

       [ 18 ! / 5! 13! ]

( 55 x 35 ) /  8568

1925 / 8568

= 0.2246 ≈ 0.225

Answer:

It's asking you to solve the question or find N

Step-by-step explanation:

In the given scenario, when the shoe company makes 360 black shoes, they will also make 270 blue shoes as the ratio was provided as 3:4.

The problem involves the concept of ratio proportion. Given the ratio of blue shoes to black shoes is 3:4, this means for every 3 blue shoes, 4 black shoes are made. If the company makes 360 black shoes, this is like 90 sets of 4 black shoes (360/4). That means 90 sets of 3 blue shoes must also be made. Therefore, the company will make 270 (90*3) blue shoes when 360 black shoes are made.

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The volume of the glass bead formed by a rectangular prism with a smaller prism removed can be found by subtracting the volume of the smaller prism from the larger prism. As an example, a larger prism of volume 60 cubic cm and smaller prism of volume 8 cubic cm gives a bead of volume 52 cubic cm.

The volume of the glass bead formed by a rectangular prism with a smaller rectangular prism removed can be calculated using the formula for the volume of a rectangular prism, which is length × width × height. The volume of the glass bead would then be the volume of the larger prism minus the volume of the smaller prism.

As an example, if the larger prism has a length of 5cm, width of 4cm, and height of 3cm, its volume would be 5cm × 4cm × 3cm = 60 cubic cm. If the smaller prism removed has a length of 2 cm, width of 2 cm, and height of 2 cm, its volume would be 2cm × 2cm × 2cm = 8 cubic cm. Subtracting the volume of the smaller prism from the larger one gives a volume of 60 cubic cm - 8 cubic cm = 52 cubic cm which is the volume of the glass bead.

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The volume of the glass bead is then 200 cm³ - 6 cm³ = 194 cm³.

Calculating the Volume of a Glass Bead

To find the volume of a glass bead with the shape of a rectangular prism with a smaller rectangular prism removed, follow these steps:

For example, if the dimensions of the larger rectangular prism are 10 cm (length), 5 cm (width), and 4 cm (height), its volume will be 200 cm³. If the smaller removed prism has dimensions of 3 cm (length), 2 cm (width), and 1 cm (height), its volume will be 6 cm³. The volume of the glass bead is then 200 cm³ - 6 cm³ = 194 cm³.

Answer: M≤ $15

Step-by-step explanation:

The answer is [ m ≤ $15 ]

The question states, "If you have no more than $15." This means that you can only have up to $15 and no more.

So, the solutions to the set are all real numbers except the numbers after 15.

Let's say you had $9. If we were to substitute 9 with m in the inequality, we would get; 9 ≤ 15. This satisfies the inequality since the number is less than 15.

Best of Luck!

Answer:

[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]

[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Step-by-step explanation:

Data given:

65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8

The sample mean would be given by:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

We can calculate the sample deviation with this formula:

[tex]s = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

And we got:

s=5.437 represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=22 the sample size

Confidence=99% or 0.99

The confidence interval for the population variance is given by:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

The degrees of freedom given by:

[tex]df=n-1=22-1=21[/tex]

The Confidence is 0.99 or 99%, the value of significance is [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and the critical values are:

[tex]\chi^2_{\alpha/2}=41.402[/tex]

[tex]\chi^2_{1- \alpha/2}=8.034[/tex]

And the confidence interval would be:

[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]

[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Answer:

Would you be able to write it in english so i can help you.

Step-by-step explanation:

Answer:

The probability that the sample mean weight will be more than 262 lb is 0.0047.

Step-by-step explanation:

The random variable X can be defined as the weight of National Football League (NFL) players now.

The mean weight is, μ = 252.8 lb.

The standard deviation of the weights is, σ = 25 lb.

A random sample of n = 50 NFL players are selected.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the sample means is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the sample means is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

The sample of players selected is quite large, i.e. n = 50 > 30, so the central limit theorem can be used to approximate the distribution of sample means.

[tex]\bar X\sim N(\mu_{\bar x}=252.8,\ \sigma_{\bar x}=3.536)[/tex]

Compute the probability that the sample mean weight will be more than 262 lb as follows:

[tex]P(\bar X>262)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{262-252.8}{3.536})\\\\=P(Z>2.60)\\\\=1-P(Z<2.60)\\\\=1-0.99534\\\\=0.00466\\\\\approx 0.0047[/tex]

*Use a z-table for the probability.

Thus, the probability that the sample mean weight will be more than 262 lb is 0.0047.

To find the probability that the sample mean will be more than 262 lb, calculate the z-score using the sample mean, population mean, standard deviation, and sample size. Then, find the corresponding probability using the standard normal distribution table. Subtract the probability from 1 to get the final result, which is approximately 0.2%.

To solve this problem, we need to use the z-score formula and the standard normal distribution table. First, calculate the z-score using the formula: z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this case, x = 262 lb, μ = 252.8 lb, σ = 25 lb, and n = 50. Plug in these values and calculate the z-score. Next, find the corresponding probability using the standard normal distribution table. Look up the z-score and find the corresponding probability. The probability that the sample mean will be more than 262 lb can be found by subtracting the probability you found from 1.

Calculating the z-score:

z = (262 - 252.8) / (25 / sqrt(50)) = 2.901.

Using the standard normal distribution table, the probability corresponding to a z-score of 2.901 is approximately 0.998. Therefore, the probability that the sample mean will be more than 262 lb is approximately 1 - 0.998 = 0.002, or 0.2%.

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Answer:

Amount invested in first account = $1,000

Amount invested in second account = $29,000

Step-by-step explanation:

Jina had total amount of $30,000 to invest last year.

Let x be the amount that Jina invested in the first account at interest rate of 9%

Mathematically,

0.09x

she invested the remaining amount in the second account at an interest rate of 5%

Mathematically,

0.05(30,000 - x)

Jina received $1540 in interest.

0.09x + 0.05(30,000 - x) = 1540

0.09x + 1500 - 0.05x = 1540

0.04x = 1540 - 1500

0.04x = 40

x = 40/0.04

x = $1,000

Therefore, Jina invested an amount of $1,000 in the first account at interest rate of 9%  

The remaining amount that she invested in the second account is

Amount invested in second account = $30,000 - $1,000

Amount invested in second account = $29,000

Therefore, Jina invested an amount of $29,000 in the second account at interest rate of 5%

Verification:

0.09x + 0.05(30,000 - x) = 1540

0.09(1000) + 0.05(30,000 - 1000) = 1540

90 + 1450 = 1540

1540 = 1540  (satisfied)

Answer:

[tex]x=-\frac{17}{8}+\frac{\sqrt{97} }{8}[/tex] or [tex]x=-\frac{17}{8}-\frac{\sqrt{97} }{8}[/tex]

Step-by-step explanation:

[tex]x^2+17x+12=-3x^2[/tex]

Add [tex]3x^2[/tex] on both sides.

[tex]x^2+17x+12+3x^2=-3x^2+3x^2[/tex]

[tex]x^2+17x+12+3x^2=0[/tex]

Combine like terms;

[tex]4x^2+17x+12=0[/tex]

Use the quadratic formula;

a=4

b=17

c=12

[tex]x=\frac{-b\frac{+}{}\sqrt{b^2-4ac} }{2a}[/tex]

[tex]x=\frac{-(17)\frac{+}{}\sqrt{(17)^2-4(4)(12)} }{2(4)}[/tex]

[tex]x=\frac{-17\frac{+}{}\sqrt{289-192} }{8}[/tex]

[tex]x=\frac{-17\frac{+}{}\sqrt{97} }{8}[/tex]

[tex]x=-\frac{17}{8}+\frac{\sqrt{97} }{8}[/tex] or [tex]x=-\frac{17}{8}-\frac{\sqrt{97} }{8}[/tex]

Shouldn't the equation read " (x - 4)² = 81 ", instead of "(1 - 4)² = 81" ?

If so, then the solutions are  x = -5  and  x = 13 .

If it's really "(1 - 4)² = 81", then that's not even an equation, and there's no solution.

The solutions to the given equation are x = -5 and x = 13.

The correct solutions to the given equation are x = -5 and x = 13.

To solve the equation (1 - 4)2 = 81:

Answer:

  9

Step-by-step explanation:

If 24 points is half the number of points all other players scored, then the points scored by all other players total 24·2 = 48.

All points together total ...

  24 + 48 = 72

A point is awarded for each game, so there were a total of 72 games.

N players will play a total of (N/2)(N-1) games if they play each opponent once. Here, each opponent is played twice, so the total number of games played by N players is ...

  N(N-1) = 72

  9·8 = 72   ⇒   N = 9

The number of players in the championship was 9.

Step-by-step explanation:

9 + x - 7

Solving like terms

x + 2

If we find the value of x

X + 2 = 0

x = - 2

Answer:

Easy! The cool thing about correlations is you can easily determine them by reading your graph, no hard brain work involved!

It should be A.

Step-by-step explanation:

By looking at the graph, you can already determine that the correlation is negative, since it's going down, not up.

Now, you need to read what is happening on the graph. As the price (X-Axis) is increasing, less people are spending their money, presumably because it's not priced affordably. So as you can see according to your Y-Axis, the amount of people buying is lowered.

Hopefully this isn't confusing!

Answer:

There are 1,585,080 ways for her to select students and assign them to the jobs

Step-by-step explanation:

The order in which the students are selected is important, since different orderings means different jobs for each student selected. So the permutations formula is used to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

[tex]P_{(n,x)} = \frac{n!}{(n-x)!}[/tex]

In this problem:

4 students selected from a set of 37. So

[tex]P_{(37,4)} = \frac{37!}{(37-4)!} = 1585080[/tex]

There are 1,585,080 ways for her to select students and assign them to the jobs

Answer:

D. x + 8

Step-by-step explanation:

factor. Find factors of x² and 48, in which, when combined, will equal 2x:

x² + 2x - 48

x               8

x              -6

(x + 8)(x - 6)

D. x + 8 is your answer.

~

Answer:

10:00 AM

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given

Initially Plane is flying at the speed of [tex]180\ km/hr[/tex]

to the [tex]15^{\circ}[/tex] North of east

Now wind started Blowing and plane started moving towards east with speed [tex]150\ km/hr[/tex]

suppose [tex]1v_o[/tex] is the speed of wind

So,

[tex]\vec{v_2}=\vec{v_1}-\vec{v_o}[/tex]

[tex]150\hat{i}=180[\cos 15\hat{i}+\sin 15\hat{j}]-\vec{v_o}[/tex]

[tex]\vec{v_o}=\hat{i}[180\cos 15-150]+\hat{j}[180\sin 15][/tex]

[tex]\vec{v_o}=\hat{i}[173.866-150]+46.58\hat{j}[/tex]

[tex]\vec{v_o}=23.86\hat{i}+46.58\hat{j}[/tex]

So magnitude of wind is

[tex]\mid v_o\mid=\sqrt{23.86^2+46.58^2}[/tex]

[tex]\mid v_o\mid=\sqrt{2738.996}[/tex]

[tex]\mid v_o\mid=52.33\ km/hr[/tex]

direction [tex]\tan \theta=\frac{46.58}{23.86}[/tex]

[tex]\theta =62.87^{\circ}[/tex] North of east

Answer:

The speed of the wind is 52.3 km/h

The direction of the wind is  243 degrees.

this is the right answer

Step-by-step explanation:

Answer:

[tex]z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123[/tex]  

The p value for a left tailed test would be:  

[tex]p_v =P(z<-1.123)=0.131[/tex]  

Since the p value is very higher we can conclude that the true proportion  of teenagers who floss twice a day is NOT less than 40%.

Step-by-step explanation:

Information given

n=400 represent the random sample given

X=149 represent the floss twice a day

[tex]\hat p=\frac{149}{400}=0.3725[/tex] estimated proportion of floss twice a day

[tex]p_o=0.4[/tex] is the value the proportion that we want to check

z would represent the statistic

[tex]p_v[/tex] represent the p value

System of hypothesis

We want to check proportion of teenagers who floss twice a day is less than 40%, so then the system of hypothesis are.:  

Null hypothesis:[tex]p \geq 0.4[/tex]  

Alternative hypothesis:[tex]p < 0.4[/tex]  

For the one sample proportion test the statistic is given by:

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

If we replace the info given we got:

[tex]z=\frac{0.3725 -0.4}{\sqrt{\frac{0.4(1-0.4)}{400}}}=-1.123[/tex]  

The p value for a left tailed test would be:  

[tex]p_v =P(z<-1.123)=0.131[/tex]  

Since the p value is very higher we can conclude that the true proportion  of teenagers who floss twice a day is NOT less than 40%.

Answer:

If its the sum then, 0+0 if its an quotient, 0 divided by 0 if its a multiply problem, 0 times 0.

Step-by-step explanation:

0 can only be the answer of a problem if its used on itself or if its multiplies by another.

Answer:

There is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.

Step-by-step explanation:

This is an hypothesis test for the lifetime of a certain ype of light bulb.

The population distribution is normal, with mean of 1,000 hours and STD of 110 hours.

The sample size for this test is n=10.

The significance level is assumed to be 0.05.

In this case, when the claim is that the new light bulb model has a longer average lifetime, so this is a right-tailed test.

For a significance level, the critical value (zc) that is bound of the rejection region is:

[tex]P(z>z_c)=0.05[/tex]

This value of zc is zc=1.645.

This value, for a sample with size n=10 is:

[tex]z_c=\dfrac{X_c-\mu}{\sigma/\sqrt{n}}\\\\\\X_c=\mu+\dfrac{z_c\cdort\sigma}{\sqrt{n}}=1000+\dfrac{1.645*110}{\sqrt{10}}=1000+57.22=1057.22[/tex]

That means that if the sample mean (of a sample of size n=10) is bigger than 1057.22, the null hypothesis will be rejected.

The Type II error happens when a false null hypothesis failed to be rejected.

We now know that the true mean of the lifetime is 1130, the probability of not rejecting the null hypothesis (H0: μ=1100) is the probability of getting a sample mean smaller than 1057.22.

The probability of getting a sample smaller than 1057.22 when the true mean is 1130 is:

[tex]z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{1057.22-1130}{110/\sqrt{10}}=\dfrac{-72.78}{34.7851}=-2.0923 \\\\\\P(M<1057.22)=P(z<-2.0923)=0.01821[/tex]

Then, there is a probability of P=0.02 of making a Type II error if the true mean is μ=1130.

Using the normal distribution and the central limit theorem, it is found that there is a 0.0001 = 0.01% probability of a type II error.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

We test if the average lifetime is longer, and a Type II error is concluding that it is not longer when in fact it is longer, hence, it is the probability of finding a sample mean below 1000 hours, which is the p-value of Z when X = 1000.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{1000 - 1130}{\frac{110}{\sqrt{10}}}[/tex]

[tex]Z = -3.74[/tex]

[tex]Z = -3.74[/tex] has a p-value of 0.0001.

0.0001 = 0.01% probability of a type II error.

A similar problem is given at https://brainly.com/question/15186499

first find the greatest possible length:

9 + 6 > x

15 > x

x < 15

Then find the lowest possible length:

x + 6 > 9

x > 9 - 6

x > 3

3 < x < 15

To determine the range of possible lengths of the third side of a sail, use the triangle inequality theorem. The range is 3 feet to 15 feet.

To determine the range of possible lengths of the third side of the sail, we can use the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

In this case, the two given sides are 9 feet and 6 feet. So, the third side must be less than the sum of these two sides and greater than the difference between these two sides.

Therefore, the range of possible lengths of the third side of the sail is 3 feet to 15 feet.

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